- #1
kingwinner
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Suppose that normal derivative = [itex]\nabla[/itex]g . n = dg/dn,
then f [itex]\nabla[/itex]g . n = f dg/dn
[I used . for dot product]
But how is this possible?
For f [itex]\nabla[/itex]g . n, I would interpret it as (f [itex]\nabla[/itex]g) . n
But f dg/dn = f ([itex]\nabla[/itex]g . n) which is DIFFERENT (note the location of brackets)
I recall that if u,v are vectors, a is a scalar constant, then (au) . v=a(u . v) = u . (av), but in our case, f is a FUNCTION! So this rule cannot be applied.
f [itex]\nabla[/itex]g . n = f dg/dn seems to be suggesting that f is a scalar constant, but consider the following case:
Let div F=[itex]\nabla[/itex] . F
If f were really a scalar constant, then by the rule, div (f [itex]\nabla[/itex]g) = [itex]\nabla[/itex] . (f [itex]\nabla[/itex]g) = f ([itex]\nabla[/itex] . ([itex]\nabla[/itex]g)) which is WRONG becuase we know that in general div(fG)=f div G + ([itex]\nabla[/itex]f) . G where f is function and G is vector field. But if this is wrong, then HOW can you justify that (f [itex]\nabla[/itex]g) . n = f ([itex]\nabla[/itex]g . n)?
I seriously can't understand this...there are two seemingly contradicting ideas crashing in my mind...
Would someone be nice enough to clear my doubts? Thanks!
then f [itex]\nabla[/itex]g . n = f dg/dn
[I used . for dot product]
But how is this possible?
For f [itex]\nabla[/itex]g . n, I would interpret it as (f [itex]\nabla[/itex]g) . n
But f dg/dn = f ([itex]\nabla[/itex]g . n) which is DIFFERENT (note the location of brackets)
I recall that if u,v are vectors, a is a scalar constant, then (au) . v=a(u . v) = u . (av), but in our case, f is a FUNCTION! So this rule cannot be applied.
f [itex]\nabla[/itex]g . n = f dg/dn seems to be suggesting that f is a scalar constant, but consider the following case:
Let div F=[itex]\nabla[/itex] . F
If f were really a scalar constant, then by the rule, div (f [itex]\nabla[/itex]g) = [itex]\nabla[/itex] . (f [itex]\nabla[/itex]g) = f ([itex]\nabla[/itex] . ([itex]\nabla[/itex]g)) which is WRONG becuase we know that in general div(fG)=f div G + ([itex]\nabla[/itex]f) . G where f is function and G is vector field. But if this is wrong, then HOW can you justify that (f [itex]\nabla[/itex]g) . n = f ([itex]\nabla[/itex]g . n)?
I seriously can't understand this...there are two seemingly contradicting ideas crashing in my mind...
Would someone be nice enough to clear my doubts? Thanks!
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